The Simon annihilator is an (n−1)-dimensional binary subspace

Choose the least coordinate $j<n$ with $s_j=1$, which exists by [F1]. For each $i\mathrm{\ne }j$, let $u^{(i)}$ have coordinate $1$ at $i$, coordinate $s_i$ at $j$, and zero elsewhere. Then $s\cdot u^{(i)}=s_i\oplus (s_j\wedge s_i)=s_i\oplus s_i=0$, so every $u^{(i)}$ belongs to $s^{\perp }$. When $n=1$, the index set $i\mathrm{\ne }j$ is empty.

For any $y\in s^{\perp }$, the equation $s\cdot y=0$ and $s_j=1$ give $y_j={⨁}_{i\mathrm{\ne }j}(s_i\wedge y_i)$; hence $y={\sum }_{i\mathrm{\ne }j}{y}_i{u}^{(i)}$, checking each coordinate. Conversely, distributing [F4] over the finite sum gives $s\cdot (\sum c_i{u}^{(i)})=\sum c_i(s\cdot u^{(i)})=0$, so the span equals $s^{\perp }$ and makes it a subspace by [F5] and [F7]. For any injective finite list $b_t=u^{(i_t)}$ from this set, the indices $i_t\mathrm{\ne }j$ are distinct, and coordinate $i_t$ of ${\sum }_r{c}_r{b}_r$ is exactly $c_t$. Thus a zero combination forces every coefficient to vanish; when the list is empty the condition is vacuous. The set is independent by [F6] and spans by [F7].

The independent spanning set from step 2.1 is a basis by [F8], and has exactly $n-1=d$ elements, so $\dim s^{\perp }=d$. Its coefficient map is bijective by the coordinate formula and independence, so $\mathrm{\mid }{s}^{\perp }\mathrm{\mid }=2^d$ because there are $2^d$ binary coefficient lists. For $n=1$, the empty list gives only $0$, and the same count is $2^0=1$.